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Of course!! So simple! &nbsp;:-)<br />

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InteractiveMind,<br />
<br />
snoyes jw gave an ingenious method. &nbsp;Perhaps you would like to know another method.<br />
<br />
Can you differentiate? &nbsp;(If not, ignore this comment entirely.) &nbsp;We're looking for an extremum of the parabola, in other words, a place where the slope is zero. &nbsp;So the first derivative should equal zero.<br />
Let f(x) = -2x^2 + 5x + 7<br />
Then f'(x) = -4x + 5<br />
Now we can set this equal to zero and solve for x.<br />
<br />
mathbiol
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InteractiveMind,

snoyes jw gave an ingenious method.  Perhaps you would like to know another method.

Can you differentiate?  (If not, ignore this comment entirely.)  We're looking for an extremum of the parabola, in other words, a place where the slope is zero.  So the first derivative should equal zero.
Let f(x) = -2x^2 + 5x + 7
Then f'(x) = -4x + 5
Now we can set this equal to zero and solve for x.

mathbiol

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Hi..<br />
<br />
I have this quadratic equation:<br />
<br />
&nbsp; &nbsp;-2x^2 + 5x + 7 = 0<br />
<br />
With the help of you lot, I've solved 'x', like so:<br />
<br />
&nbsp; &nbsp;x = -1<br />
&nbsp; &nbsp;or<br />
&nbsp; &nbsp;x = 7 / 2<br />
<br />
Now, I need to find the *highest*point of the graph (in terms of the y-axis)... Do I need to use trial and error for this? Or can it be solved via algebra?<br />
<br />
Regards;<br />

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Hi..

I have this quadratic equation:

   -2x^2 + 5x + 7 = 0

With the help of you lot, I've solved 'x', like so:

   x = -1
   or
   x = 7 / 2

Now, I need to find the *highest*point of the graph (in terms of the y-axis)... Do I need to use trial and error for this? Or can it be solved via algebra?

Regards;


Get highest y-axis point of quadratic graph

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